Question

Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 m to the right of antenna \(A\). Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna \(B\). The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q\)? (b) What is the longest wavelength for which there will be constructive interference at point \(Q\)?

Short answer

The longest wavelength for destructive interference is 240 m, and for constructive interference is 120 m.

Step-by-step solution

Understanding the Problem

We need to determine the conditions for both destructive and constructive interference at the point Q on the line that extends through the two radio antennas, A and B. Given the distances, antenna B is 120 m from antenna A, and point Q is 40 m from antenna B.

Calculate the Path Difference

The total distance from antenna A to point Q is 120 m + 40 m = 160 m. The distance from antenna B to point Q is 40 m. The path difference, \( \Delta d \), is the difference between these two distances: \( \Delta d = 160 \text{ m} - 40 \text{ m} = 120 \text{ m} \).

Determine Condition for Destructive Interference

For destructive interference, the path difference should be an odd multiple of half the wavelength, i.e., \( \Delta d = \left( n + \frac{1}{2} \right) \lambda \), where \( n \) is an integer (0, 1, 2, ...). The longest wavelength occurs when \( n = 0 \). Setting \( n = 0 \) gives: \( 120 = \frac{1}{2} \lambda \), so \( \lambda = 240 \) m.

Determine Condition for Constructive Interference

For constructive interference, the path difference should be a multiple of the wavelength, i.e., \( \Delta d = m \lambda \), where \( m \) is an integer. The longest wavelength occurs when \( m = 1 \). Using \( \Delta d = 120 \) m gives: \( 120 = 1 \cdot \lambda \), so \( \lambda = 120 \) m.

Key concepts

Destructive Interference

When waves meet, their effects can either cancel each other out or amplify. Destructive interference happens when the peak of one wave meets the trough of another, leading them to cancel.
The result is often a reduction in wave amplitude or, in the case of identical waves, complete cancellation.
For destructive interference to occur, the path difference (\(\Delta d\)) between two waves must be an odd multiple of half the wavelength (\(\lambda\)).

• The formula for the condition is \[\Delta d = \left(n + \frac{1}{2}\right) \lambda\], where \(n\) is an integer (0, 1, 2, ...).
• The longest wavelength for destructive interference is when \(n = 0\), simplifying the formula to \(\Delta d = \frac{\lambda}{2}\).

In our problem, \(\Delta d\) is 120 m, leading to \(\lambda = 240\) m when \(n = 0\).This means point \(Q\) experiences destructive interference at this wavelength.

Constructive Interference

Constructive interference occurs when waves meet in such a way that their crests and troughs align.
This alignment leads the wave amplitudes to add together, resulting in a stronger and more pronounced wave.
The condition for constructive interference is met when the path difference (\(\Delta d\)) is a whole number multiple of the wavelength (\(\lambda\)).

• Expressed mathematically as \[\Delta d = m \lambda\], where \(m\) is an integer (1, 2, 3, ...).
• The longest wavelength occurs when \(m = 1\), as it represents the fundamental mode of interference.

In our scenario, with \(\Delta d = 120\) m, the longest wavelength for constructive interference is \(\lambda = 120\) m, making point \(Q\) a zone of amplification at this wavelength.

Path Difference

Path difference is critical in understanding wave interference. It refers to the difference in distance traveled by two waves from their sources to a common point.
In terms of interference, this difference determines whether waves will constructively or destructively interfere at that point.
The concept can be explained as follows:

• For constructive interference, the path difference should be a multiple of the wavelength: \(m \lambda\).
• For destructive interference, it should be an odd multiple of half the wavelength: \(\left(n+\frac{1}{2}\right) \lambda\).

In the given exercise, the path difference from antennas \(A\) and \(B\) to point \(Q\) is calculated as 120 m (160 m - 40 m).
This path difference guides us in determining the conditions for both constructive and destructive interference.

Wavelength Calculation

Calculating the wavelength is key to understanding interference patterns. Wavelength (\(\lambda\)) can be found using the interference conditions based on the path difference.
The method relies on understanding how the path difference relates to \(\lambda\):

• For constructive interference, use the formula \(\Delta d = m \lambda\)
• For destructive interference, use \(\Delta d = \left(n + \frac{1}{2}\right) \lambda\)

By plugging the path difference and the required condition's integer (\(m\) or \(n\)) into these formulas, we can solve for \(\lambda\).
In our problem, the given path difference of 120 m helps determine wavelengths for both types of interference:

• For constructive interference: \(\lambda = 120\) m (\(m = 1\)
• For destructive interference: \(\lambda = 240\) m (\(n = 0\)

This comprehensive understanding assists in visualizing how waves interact at different points.